Problems

2003 Paper 3 Q7

D: 1700.0 B: 1484.0

conics perpendicular locus geometric proof coordinate geometry line equations circle right angle

In the $x$--$y$ plane, the point $A$ has coordinates $(a\,,0)$ and the point $B$ has coordinates $(0\,,b)\,$, where $a$ and $b$ are positive. The point $P\,$, which is distinct from $A$ and $B$, has coordinates~$(s,t)\,$. $X$ and $Y$ are the feet of the perpendiculars from $P$ to the $x$--axis and $y$--axis respectively, and $N$ is the foot of the perpendicular from $P$ to the line $AB\,$. Show that the coordinates $(x\,,y)$ of $N$ are given by \[ x= \frac {ab^2 -a(bt-as)}{a^2+b^2} \;, \ \ \ y = \frac{a^2b +b(bt-as)}{a^2+b^2} \;. \] Show that, if $\ds \ \left( \frac{t-b} s\right)\left( \frac t {s-a}\right) = -1\;$, then $N$ lies on the line $XY\,$. Give a geometrical interpretation of this result.

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2000 Paper 3 Q1

D: 1700.0 B: 1485.5

coordinate geometry rectangular hyperbola conics curve sketching tangent-normal symmetry geometric proof locus

Sketch on the same axes the two curves $C_1$ and $C_2$, given by

\begin{align*} C_1: && x y & = 1 \\ C_2: && x^2-y^2 & = 2 \end{align*}

The curves intersect at $P$ and $Q$. Given that the coordinates of $P$ are $(a,b)$ (which you need not evaluate), write down the coordinates of $Q$ in terms of $a$ and $b$. The tangent to $C_1$ through $P$ meets the tangent to $C_2$ through $Q$ at the point $M$, and the tangent to $C_2$ through $P$ meets the tangent to $C_1$ through $Q$ at $N$. Show that the coordinates of $M$ are $(-b,a)$ and write down the coordinates of $N$. Show that $PMQN$ is a square.

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1997 Paper 3 Q8

D: 1700.0 B: 1484.0

linear transformations rotation matrix diagonalisation symmetric matrix ellipse quadratic form matrix multiplication conics optimisation

Let $R_{\alpha}$ be the $2\times2$ matrix that represents a rotation through the angle $\alpha$ and let $$A=\begin{pmatrix}a&b\\b&c\end{pmatrix}.$$

Find in terms of $a$, $b$ and $c$ an angle $\alpha$ such that $R_{-\alpha}AR_{\alpha}$ is a diagonal matrix (i.e. has the value zero in top-right and bottom-left positions).
Find values of $a$, $b$ and $c$ such that the equation of the ellipse \[x^2+(y+2x\cot2\theta)^2=1\qquad(0 < \theta < \tfrac{1}{4}\pi)\] can be expressed in the form \[\begin{pmatrix}x&y\end{pmatrix}A\begin{pmatrix}x\\y\end{pmatrix}=1.\] Show that, for this $A$, $R_{-\alpha}AR_{\alpha}$ is diagonal if $\alpha=\theta$. Express the non--zero elements of this matrix in terms of $\theta$.
Deduce, or show otherwise, that the minimum and maximum distances from the centre to the circumference of this ellipse are $\tan\theta$ and $\cot\theta$.

Solution: \begin{questionparts} \item \begin{align*} R_{-\alpha}AR_{\alpha} &= \begin{pmatrix} \cos \alpha & \sin\alpha \\ -\sin \alpha & \cos \alpha \end{pmatrix}\begin{pmatrix} a & b \\ b & c \end{pmatrix} \begin{pmatrix} \cos \alpha & -\sin\alpha \\ \sin \alpha & \cos \alpha \end{pmatrix} \\ &= \begin{pmatrix} \cos \alpha & \sin\alpha \\ -\sin \alpha & \cos \alpha \end{pmatrix} \begin{pmatrix} a\cos \alpha + b \sin \alpha & -a\sin\alpha + b \cos\alpha \\ b\cos\alpha + c \sin\alpha & c\cos\alpha-b\sin\alpha \end{pmatrix} \\ &= \begin{pmatrix} a\cos^2\alpha+2b\sin\alpha\cos\alpha+c\sin^2\alpha & -a\sin\alpha\cos \alpha+b\cos^2\alpha +c\sin\alpha\cos\alpha-b\sin^2 \alpha\\ (c-a)\sin\alpha\cos \alpha +b(\cos^2\alpha-\sin^2 \alpha) & a\sin^2 \alpha -2b\sin\alpha\cos\alpha+c\cos^2\alpha \end{pmatrix} \\ &= \begin{pmatrix} * & \frac{c-a}{2}\sin2\alpha+b \cos 2\alpha\\\frac{c-a}{2}\sin2\alpha+b \cos 2\alpha & * \end{pmatrix} \end{align*} Therefore this will be diagonal if $\tan 2\alpha = \frac{2b}{a-c} \Rightarrow \alpha = \frac12 \tan^{-1} \l \frac{2b}{a-c} \r$ \item \begin{align*} x^2+(y+2x\cot2\theta)^2 &= x^2(1 + 4\cot^22\theta) + 4\cot2\theta xy + y^2 \\ &= \begin{pmatrix}x&y\end{pmatrix}\begin{pmatrix} 1 + 4\cot^22\theta & 2\cot 2\theta \\ 2\cot 2\theta & 1 \end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix} \end{align*} Plugging this $\mathbf{A}$ in our result from before we discover \begin{align*} \frac12 \tan^{-1} \l \frac{2b}{a-c} \r &= \frac12 \tan^{-1} \l \frac{4\cot 2\theta}{1 + 4\cot^22\theta-1} \r \\ &= \frac12 \tan^{-1} \l \tan 2 \theta \r \\ &= \theta \end{align*} Therefore, the matrix will be: \begin{align*} & \textrm{diag}\begin{pmatrix} (1+4\cot^2 2\theta)\cos^2 \theta + 4\cot2\theta \sin\theta\cos\theta + \sin^2 \theta \\ (1+4\cot^2 2\theta)\sin^2 \theta - 4\cot2\theta \sin\theta\cos\theta + \cos^2 \theta \end{pmatrix} \\ =& \textrm{diag}\begin{pmatrix} \cos^2\theta + \frac{\cos^2 2\theta}{\sin^2 \theta} + 2\cos 2\theta + \sin^2 \theta \\ \sin^2\theta + \frac{\cos^2 2\theta}{\cos^2 \theta} - 2\cos 2\theta + \cos^2 \theta \end{pmatrix} \\ =& \textrm{diag}\begin{pmatrix} 1 + \cos 2\theta \l \frac{\cos2\theta}{\sin^2 \theta} + 2\r \\ 1 + \cos 2\theta \l \frac{\cos2\theta}{\cos^2 \theta} - 2\r \\ \end{pmatrix} \\ =& \textrm{diag}\begin{pmatrix} 1 + \cos 2\theta \l \frac{\cos^2 \theta + \sin^2 \theta}{\sin^2 \theta}\r \\ 1 -\cos 2\theta \l \frac{-\cos^2 \theta - \sin^2 \theta}{\cos^2 \theta}\r \\ \end{pmatrix} \\ =& \textrm{diag}\begin{pmatrix} 1 + (\cos^2\theta - \sin^2 \theta) \cosec^2 \theta \\ 1 - (\cos^2\theta - \sin^2 \theta) \sec^2 \theta \\ \end{pmatrix} \\ =& \textrm{diag}\begin{pmatrix} \cot^2 \theta \\ \tan^2 \theta \\ \end{pmatrix} \\ \end{align*} Therefore this is a rotation of an ellipse with equation: $(\cot \theta x)^2 + (\tan \theta y)^2 = 1$, ie the shortest side and longest side are $\cot \theta$ and $\tan \theta$ respectively, but we know since $0 < \theta < \tfrac{1}{4}\pi$ the shortest will be $\tan \theta$ and the longest $\cot \theta$.

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1994 Paper 1 Q5

D: 1500.0 B: 1516.0

conics parabola locus midpoint tangent perpendicular curve sketching parametric

A parabola has the equation $y=x^{2}.$ The points $P$ and $Q$ with coordinates $(p,p^{2})$ and $(q,q^{2})$ respectively move on the parabola in such a way that $\angle POQ$ is always a right angle.

Find and sketch the locus of the midpoint $R$ of the chord $PQ.$
Find and sketch the locus of the point $T$ where the tangents to the parabola at $P$ and $Q$ intersect.

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1992 Paper 3 Q9

D: 1700.0 B: 1515.1

conics ellipse polar coordinates tangent-normal focus-directrix parametric trigonometric diameter

The straight line $OSA,$ where $O$ is the origin, bisects the angle between the positive $x$ and $y$ axes. The ellipse $E$ has $S$ as focus. In polar coordinates with $S$ as pole and $SA$ as the initial line, $E$ has equation $\ell=r(1+e\cos\theta).$ Show that, at the point on $E$ given by $\theta=\alpha,$ the gradient of the tangent to the ellipse is given by \[ \frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\sin\alpha-\cos\alpha-e}{\sin\alpha+\cos\alpha+e}. \] The points on $E$ given by $\theta=\alpha$ and $\theta=\beta$ are the ends of a diameter of $E$. Show that \[ \tan(\alpha/2)\tan(\beta/2)=-\frac{1+e}{1-e}. \] [Hint. A diameter of an ellipse is a chord through its centre.]

Solution:

\begin{align*} && \ell &= r(1 + e \cos \theta) \\ \Rightarrow && 0 &= \frac{\d r}{\d \theta}(1 + e \cos \theta) - re \sin \theta \\ \Rightarrow && \frac{\d r}{\d \theta} &= \frac{re \sin \theta}{1+e \cos \theta} \end{align*} Suppose we consider the $(x',y')$ plane, which is essentially the $x-y$ plan rotated by $45^\circ$, then we would have \begin{align*} && \frac{\d y'}{ \d x'} &= \frac{\frac{\d y'}{\d \theta}}{\frac{\d x'}{\d \theta}} \\ &&&= \frac{\frac{\d r}{\d \theta} \sin \theta + r\cos \theta}{\frac{\d r}{\d \theta} \cos \theta - r\sin \theta} \\ &&&= \frac{\frac{re \sin \theta}{1+e \cos \theta} \sin \theta + r\cos \theta}{\frac{re \sin \theta}{1+e \cos \theta} \cos\theta -r\sin \theta} \\ &&&= \frac{re\sin^2 \theta+r \cos \theta(1+e \cos \theta)}{re\sin \theta \cos \theta -r \sin \theta (1+e \cos \theta)} \\ &&&= \frac{\cos \theta + e \cos^2 \theta+e \sin^2 \theta}{-\sin \theta} \\ &&&= \frac{\cos \theta + e}{-\sin \theta} \end{align*} Since our frame is rotated by $45^\circ$ we need to consider the appropriate gradient for this. We know that $m = \tan \theta$ so $m' = \tan (\theta+45^{\circ}) = \frac{1+m}{1-m}$ therefore we should have \begin{align*} && \frac{\d y}{ \d x} &= \frac{1+\frac{\cos \theta + e}{-\sin \theta}}{1-\frac{\cos \theta + e}{-\sin \theta}} \\ &&&= \frac{\cos \theta - \sin \theta + e}{-\sin \theta - \cos \theta-e} \\ &&&= \frac{\sin \theta - \cos \theta -e}{\sin \theta + \cos \theta +e} \end{align*} As required. The tangents at those points are parallel, therefore \begin{align*} && \frac{\cos \alpha+e}{\sin \alpha} &= \frac{\cos \beta+e}{\sin \beta} \\ \Rightarrow && \frac{\frac{1-\tan^2 \frac{\alpha}{2}}{1+\tan^2 \frac{\alpha}{2}}+e}{\frac{2\tan \frac{\alpha}{2}}{1+\tan^2 \frac{\alpha}{2}}} &= \frac{\frac{1-\tan^2 \frac{\beta}{2}}{1+\tan^2 \frac{\beta}{2}}+e}{\frac{2\tan \frac{\beta}{2}}{1+\tan^2 \frac{\beta}{2}}} \\ && \frac{1-\tan^2 \frac{\alpha}{2}+e(1+\tan^2\frac{\alpha}{2})}{2\tan\frac{\alpha}{2}} &= \frac{1-\tan^2 \frac{\beta}{2}+e(1+\tan^2\frac{\beta}{2})}{2\tan\frac{\beta}{2}} \\ && \frac{(1+e)+(e-1)\tan^2 \frac{\alpha}{2}}{2\tan \frac{\alpha}{2}} &= \frac{(1+e)+(e-1)\tan^2 \frac{\beta}{2}}{2\tan \frac{\beta}{2}} \\ && \frac{(1+e)}{\tan\frac{\alpha}2} - (1-e)\tan\frac{\alpha}2 &= \frac{(1+e)}{\tan\frac{\beta}2} - (1-e)\tan\frac{\beta}2 \end{align*} ie both $\tan \frac{\alpha}{2}$ and $\tan \frac{\beta}{2}$ are roots of a quadratic of the form $(1-e)x^2-cx-(1+e)$ but this means the product of the roots is $-\frac{1+e}{1-e}$

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1991 Paper 2 Q2

D: 1600.0 B: 1528.5

implicit differentiation conics hyperbola tangent-normal asymptotes angle bisector curve sketching

The equation of a hyperbola (with respect to axes which are displaced and rotated with respect to the standard axes) is \[ 3y^{2}-10xy+3x^{2}+16y-16x+15=0.\tag{$\dagger$} \] By differentiating $(\dagger)$, or otherwise, show that the equation of the tangent through the point $(s,t)$ on the curve is \[ y=\left(\frac{5t-3s+8}{3t-5s+8}\right)x-\left(\frac{8t-8s+15}{3t-5s+8}\right). \] Show that the equations of the asymptote (the limiting tangents as $s\rightarrow\infty$) are \[ y=3x-4\qquad\mbox{ and }\qquad3y=x-4. \] {[}Hint: You will need to find a relationship between $s$ and $t$ which is valid in the limit as $s\rightarrow\infty.${]} Show that the angle between one asymptote and the $x$-axis is the same as the angle between the other asymptote and the $y$-axis. Deduce the slopes of the lines that bisect the angles between the asymptotes and find the equations of the axes of the hyperbola.

Solution: \begin{align*} && 0 &= 3y^{2}-10xy+3x^{2}+16y-16x+15 \\ \Rightarrow && 0 &= 6y \frac{\d y}{\d x} - 10x \frac{\d y}{\d x} - 10y + 6x+ 16 \frac{\d y}{\d x } - 16 \\ &&&= \frac{\d y}{\d x} \left (6y - 10x +16 \right) - (10y-6x+16) \\ \Rightarrow && \frac{\d y}{\d x} &= \frac{5y-3x+8}{3y-5x+8} \\ \Rightarrow && \frac{y-t}{x-s} &= \frac{5t-3s+8}{3t-5s+8} \\ && y &= \left(\frac{5t-3s+8}{3t-5s+8}\right)x -\left(\frac{5t-3s+8}{3t-5s+8}\right)s + t \\ &&&= \left(\frac{5t-3s+8}{3t-5s+8}\right)x - \frac{(5ts-3s^2+8s)-(3t^2-5st+8t)}{3t-5s+8} \\ &&&= \left(\frac{5t-3s+8}{3t-5s+8}\right)x - \frac{(5ts-3s^2+8s)-(3t^2-5st+8t)}{3t-5s+8} \\ &&&= \left(\frac{5t-3s+8}{3t-5s+8}\right)x - \frac{8s-8t-(3s^2+3t^2-10st)}{3t-5s+8} \\ &&&= \left(\frac{5t-3s+8}{3t-5s+8}\right)x - \frac{8s-8t-(16s-16t-15)}{3t-5s+8} \\ &&&= \left(\frac{5t-3s+8}{3t-5s+8}\right)x - \frac{8t-8s+15}{3t-5s+8} \\ \end{align*} While $x \to \infty$ we still have $3 \frac{y^2}{x^2} - 10 \frac{y}{x} + 3 + 16 \frac{y}{x^2} - 16\frac{1}{x} + 15 \frac{1}{x^2} = 0$, ie if $\frac{y}{x} = k$, then $3k^2 - 10k + 3 \to 0 \Rightarrow k \to 3, \frac13$. Therefore, as $s \to \infty$ we can write \begin{align*} && y &= \left(\frac{5\frac{t}{s}-3+8\frac{1}{s}}{3\frac{t}{s}-5+8\frac1{s}}\right)x - \frac{8\frac{t}s-8+15\frac{1}{s}}{3\frac{t}{s}-5+8\frac{1}{s}} \\ k \to 3: &&& \to \left(\frac{15-3}{9-5}\right)x - \frac{24-8}{9-5} \\ &&&= 3x - 4 \\ k \to \frac13: && &\to \left(\frac{\frac53-3}{1-5}\right)x - \frac{\frac83-8}{1-5} \\ &&&= \frac13 x - \frac43 \end{align*} Therefore the equations are $y = 3x-4$ and $3y=x-4$ The lines are parallel to $y = 3x$ and $y = \frac13x$, so by considering the triangles formed with the origin and a point $1$ along the $x$ or $y$ axis we can see the angles are identical. This means the line $y = x$ is parallel to one axis and $y = -x$ is parallel to the other. They must meet where our two lines meet which is $(1,-1)$, so our lines are $y = x-2$ and $y = -x$

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1991 Paper 3 Q9

D: 1700.0 B: 1485.6

polar coordinates conics ellipse parametric arc length integration binomial expansion eccentricity coordinate transformation

The parametric equations $E_{1}$ and $E_{2}$ define the same ellipse, in terms of the parameters $\theta_{1}$ and $\theta_{2}$, (though not referred to the same coordinate axes). \begin{alignat*}{2} E_{1}:\qquad & x=a\cos\theta_{1}, & \quad & y=b\sin\theta_{1},\\ E_{2}:\qquad & x=\dfrac{k\cos\theta_{2}}{1+e\cos\theta_{2}}, & \quad & y=\dfrac{k\sin\theta_{2}}{1+e\cos\theta_{2}}, \end{alignat*} where $0< b< a,$ $0< e< 1$ and $0< k$. Find the position of the axes for $E_{2}$ relative to the axes for $E_{1}$ and show that $k=a(1-e^{2})$ and $b^{2}=a^{2}(1-e^{2}).$ {[}The standard polar equation of an ellipse is $r=\dfrac{\ell}{1+e\cos\theta}.]$ By considering expressions for the length of the perimeter of the ellipse, or otherwise, prove that \[ \int_{0}^{\pi}\sqrt{1-e^{2}\cos^{2}\theta}\,\mathrm{d}\theta=\int_{0}^{\pi}\frac{1-e^{2}}{(1+e\cos\theta)^{2}}\sqrt{1+e^{2}+2e\cos\theta}\,\mathrm{d}\theta. \] Given that $e$ is so small that $e^{6}$ may be neglected, show that the value of either integral is \[ \tfrac{1}{64}\pi(64-16e^{2}-3e^{4}). \]

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1990 Paper 3 Q6

D: 1700.0 B: 1500.0

linear transformations eigenvectors eigenvalues matrix conics ellipse tangent-normal area inverse matrix

The transformation $T$ from $\begin{pmatrix} x \\ y \end{pmatrix}$ to $\begin{pmatrix} X \\ Y \end{pmatrix}$ is given by \[ \begin{pmatrix}X\\ Y \end{pmatrix}=\frac{2}{5}\begin{pmatrix}9 & -2\\ -2 & 6 \end{pmatrix}\begin{pmatrix}x\\ y \end{pmatrix}. \] Show that $T$ leaves the vector $\begin{pmatrix} 1\\ 2 \end{pmatrix}$ unchanged in direction but multiplied by a scalar, and that $\begin{pmatrix} 2\\ -1 \end{pmatrix}$ is similarly transformed. The circle $C$ whose equation is $x^{2}+y^{2}=1$ transforms under $T$ to a curve $E$. Show that $E$ has equation \[ 8X^{2}+12XY+17Y^{2}=80, \] and state the area of the region bounded by $E$. Show also that the greatest value of $X$ on $E$ is $2\sqrt{17/5}.$ Find the equation of the tangent to $E$ at the point which corresponds to the point $\frac{1}{5}(3,4)$ on $C$.

Solution: \begin{align*} T\begin{pmatrix}1\\ 2 \end{pmatrix} &= \frac{2}{5}\begin{pmatrix}9 & -2\\ -2 & 6 \end{pmatrix}\begin{pmatrix}1\\ 2 \end{pmatrix} \\ &= \frac25\begin{pmatrix}9 - 4\\ -2+12 \end{pmatrix} \\ &= \begin{pmatrix}2\\ 4 \end{pmatrix} \\ &= 2 \begin{pmatrix}1\\ 2 \end{pmatrix} \end{align*} \begin{align*} T\begin{pmatrix}1\\ 2 \end{pmatrix} &= \frac{2}{5}\begin{pmatrix}9 & -2\\ -2 & 6 \end{pmatrix}\begin{pmatrix}2\\ -1 \end{pmatrix} \\ &= \frac25\begin{pmatrix}18+2\\ -4-6 \end{pmatrix} \\ &= \begin{pmatrix}8\\ -4 \end{pmatrix} \\ &= 4 \begin{pmatrix}2\\ -1 \end{pmatrix} \end{align*} Consider $T^{-1} = \frac{5}{2} \frac{1}{50}\begin{pmatrix}6 & 2\\ 2 & 9 \end{pmatrix}$, so $T^{-1} \begin{pmatrix}X\\ Y \end{pmatrix} = \begin{pmatrix}x\\ y \end{pmatrix}$ and so: \begin{align*} x^2 + y^2 & = \begin{pmatrix}x& y \end{pmatrix}\begin{pmatrix}x\\ y \end{pmatrix} \\ &= \begin{pmatrix}X& Y \end{pmatrix} (T^{-1})^T T^{-1} \begin{pmatrix}X\\ Y \end{pmatrix} \\ &= \frac{1}{400}\begin{pmatrix}X& Y \end{pmatrix}\begin{pmatrix}6 & 2\\ 2 & 9 \end{pmatrix}\begin{pmatrix}6 & 2\\ 2 & 9 \end{pmatrix} \begin{pmatrix}X\\ Y \end{pmatrix} \\ &= \frac{1}{400}\begin{pmatrix}X& Y \end{pmatrix}\begin{pmatrix}6 & 2\\ 2 & 9 \end{pmatrix} \begin{pmatrix}6X+2Y\\ 2X+9Y \end{pmatrix} \\ &= \frac{1}{400}\begin{pmatrix}X& Y \end{pmatrix} \begin{pmatrix}6(6X+2Y)+2(2X+9Y)\\ 2(6X+2Y)+9(2X+9Y) \end{pmatrix} \\ &= \frac{1}{400}\begin{pmatrix}X& Y \end{pmatrix} \begin{pmatrix}40X+30Y\\ 30X +85Y \end{pmatrix} \\ &= \frac{1}{80}\begin{pmatrix}X& Y \end{pmatrix} \begin{pmatrix}8X+6Y\\ 6X +17Y \end{pmatrix} \\ &= \frac{1}{80} \l 8X^2 + 12XY + 17Y^2\r \end{align*} Therefore $8X^2 + 12XY + 17Y^2 = 80$. The area will be $\det T \cdot \pi = \frac{4}{25} \cdot 50 \cdot \pi = 8 \pi$. Differentiating we obtain $2 \cdot 8 \cdot X \cdot \frac{dX}{dY} + 2 \cdot 6 \cdot X + 2 \cdot 6 \cdot Y \cdot \frac{dX}{dY} + 2 \cdot 17 Y \Rightarrow \frac{dX}{dY} = -\frac{6X + 17Y}{8X+6Y}$, at a maximum (or minimum, $6X = -17Y$). Therefore \begin{align*} \Rightarrow && 8X^2 + 12 \cdot \frac{6}{17}X^2 + 17 ( -\frac{6}{17} X)^2 &= 80 \\ \Rightarrow && \frac{100}{17}X^2 &= 80 \\ \Rightarrow &&X^2 &= \frac{17 \cdot 4}{5} \\ \Rightarrow && |X| = 2 \sqrt {\frac{17}{5}} \end{align*} The point $\frac15 (3,4)$ maps to \begin{align*} \frac{2}{5}\frac{1}{5}\begin{pmatrix}9 & -2\\ -2 & 6 \end{pmatrix}\begin{pmatrix}3\\ 4 \end{pmatrix} &= \frac{2}{25} \begin{pmatrix}19\\ 18 \end{pmatrix} \end{align*} So the point is $(\frac{38}{25}, \frac{36}{25})$, with gradient $\frac{dY}{dX} = -\frac{8X+6Y}{6X + 17Y}$ which is $-\frac{8 \cdot 19+6 \cdot 18}{6\cdot 19 + 17 \cdot 18} = -\frac{13}{21}$ therefore the equation is $21Y+13X = 50$

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1989 Paper 3 Q6

D: 1700.0 B: 1500.0

polar coordinates conics tangent-normal chord equation angle bisector trigonometric identities focus-directrix parametric

Show that, for a given constant $\gamma$ $(\sin\gamma\neq0)$ and with suitable choice of the constants $A$ and $B$, the line with cartesian equation $lx+my=1$ has polar equations \[ \frac{1}{r}=A\cos\theta+B\cos(\theta-\gamma). \] The distinct points $P$ and $Q$ on the conic with polar equations \[ \frac{a}{r}=1+e\cos\theta \] correspond to $\theta=\gamma-\delta$ and $\theta=\gamma+\delta$ respectively, and $\cos\delta\neq0.$ Obtain the polar equation of the chord $PQ.$ Hence, or otherwise, obtain the equation of the tangent at the point where $\theta=\gamma.$ The tangents at $L$ and $M$ to a conic with focus $S$ meet at $T.$ Show that $ST$ bisects the angle $LSM$ and find the position of the intersection of $ST$ and $LM$ in terms of your chosen parameters for $L$ and $M.$

Solution: \begin{align*} && \frac1{r} &= A \cos \theta + B \cos (\theta - \gamma) \\ &&&= A \cos \theta + B \cos \theta \cos \gamma + B \sin \theta \sin \gamma \\ &&&= (A+B \cos \gamma) \cos \theta + B \sin \gamma \sin \theta \\ \Longleftrightarrow && 1 &= (A+B \cos \gamma) x + B \sin \gamma y \end{align*} So if we choose $B = \frac{m}{\sin \gamma}$ and $A = l-m \cot \gamma$ we have the desired result. \begin{align*} && \frac{1 + e \cos (\gamma -\delta)}a &= A \cos (\gamma - \delta) + B \cos (\gamma - \delta - \gamma) \\ &&&= A \cos(\gamma-\delta) +B \cos \delta\\ && \frac{1 + e \cos (\gamma +\delta)}{a} &= A \cos (\gamma + \delta) + B \cos (\gamma + \delta - \gamma) \\ &&&= A \cos(\gamma + \delta) + B \cos \delta\\ \Rightarrow && \frac1{r} &= \frac{e}{a} \cos \theta + \frac{1}{a \cos \delta} \cos (\theta - \gamma) \\ \lim{\delta \to 0} &&\frac1{r} &= \frac{e}{a} \cos \theta+ \frac{1}{a} \cos (\theta - \gamma) \end{align*} Suppose we have have points $L$ and $M$ with $\theta = \gamma_L, \gamma_M$ then our tangents are: \begin{align*} && \frac{a}{r} &= \cos \theta + \cos (\theta - \gamma_L) \\ && \frac{a}{r} &= \cos \theta + \cos (\theta - \gamma_M) \\ \Rightarrow && 0 &= \cos (\theta - \gamma_L) -\cos(\theta - \gamma_M) \\ &&&= - 2 \sin \frac{(\theta - \gamma_L)+(\theta - \gamma_M)}{2} \sin \frac{(\theta - \gamma_L)-(\theta - \gamma_M)}{2} \\ &&&= -2 \sin \left ( \theta - \frac{\gamma_L+\gamma_M}2 \right) \sin \left ( \frac{\gamma_M - \gamma_L}{2}\right) \\ \Rightarrow && \theta &= \frac{\gamma_L+\gamma_M}{2} \end{align*} Therefore clearly $ST$ bisects $LSM$. The line $LM$ can be seen as the chord from the points $\frac{\gamma_L+\gamma_M}{2} \pm \frac{\gamma_L-\gamma_M}{2}$, so the line is: \begin{align*} && \frac{a}{r} &= e \cos \theta + \frac{1}{\cos \left ( \frac{\gamma_L-\gamma_M}{2}\right)} \cos \left (\theta - \frac{\gamma_L+\gamma_M}{2} \right) \end{align*} and we want the point on the line where $\theta =\frac{\gamma_L+\gamma_M}{2}$ so \begin{align*} && \frac{a}{r} &= e \cos \left ( \frac{\gamma_L+\gamma_M}{2} \right) + \frac{1}{\cos \left ( \frac{\gamma_L-\gamma_M}{2}\right)} \\ \Rightarrow && r &= \frac{a}{e \cos \left ( \frac{\gamma_L+\gamma_M}{2} \right) + \frac{1}{\cos \left ( \frac{\gamma_L-\gamma_M}{2}\right)}} \end{align*}

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1988 Paper 3 Q8

D: 1700.0 B: 1500.0

conics parabola tangent-normal parametric determinant area of triangle concurrent lines

Find the equations of the tangent and normal to the parabola $y^{2}=4ax$ at the point $(at^{2},2at).$ For $i=1,2,$ and 3, let $P_{i}$ be the point $(at_{i}^{2},2at_{i}),$ where $t_{1},t_{2}$ and $t_{3}$ are all distinct. Let $A_{1}$ be the area of the triangle formed by the tangents at $P_{1},P_{2}$ and $P_{3},$ and let $A_{2}$ be the area of the triangle formed by the normals at $P_{1},P_{2}$ and $P_{3}.$ Using the fact that the area of the triangle with vertices at $(x_{1},y_{1}),(x_{2},y_{2})$ and $(x_{3},y_{3})$ is the absolute value of \[ \tfrac{1}{2}\det\begin{pmatrix}x_{1} & y_{1} & 1\\ x_{2} & y_{2} & 1\\ x_{3} & y_{3} & 1 \end{pmatrix}, \] show that $A_{3}=(t_{1}+t_{2}+t_{3})^{2}A_{1}.$ Deduce a necessary and sufficient condition in terms of $t_{1},t_{2}$ and $t_{3}$ for the normals at $P_{1},P_{2}$ and $P_{3}$ to be concurrent.

Solution: $\frac{dy}{dt} = 2a, \frac{dx}{dt} = 2at \Rightarrow \frac{dy}{dx} = \frac{1}{t}$. Therefore the equation of the tangent will be $\frac{y - 2at}{x-at^2} = \frac{1}{t} \Rightarrow y = \frac1tx +at$ and normal will be $\frac{y-2at}{x-at^2} = -t \Rightarrow y = t(at^2-x+2a)$. The tangents will meet when: \begin{align*} && \begin{cases} t_iy -x &= at_i^2 \\ t_j y - x &= at_j^2 \\ \end{cases} \\ \Rightarrow &&(t_i - t_j)y &= a(t_i-t_j)(t_i+t_j) \\ \Rightarrow && y &= a(t_i+t_j) \\ && x &= at_it_j \end{align*} The normals will meet when: \begin{align*} && \begin{cases} y +t_i x &= at_i^3+2at_i \\ y +t_j x &= at_j^3+2at_j \\ \end{cases} \\ \Rightarrow &&(t_i - t_j)x &= a(t_i-t_j)(t_i^2+t_it_j+t_j^2+2) \\ \Rightarrow && x&= a(t_i^2+t_it_j+t_j^2+2) \\ && y &= -at_it_j(t_i+t_j) \end{align*} Therefore the area of our triangles will be: \begin{align*} \tfrac{1}{2}\det\begin{pmatrix}at_1t_2 & a(t_1+t_2) & 1\\ at_2t_3 & a(t_2+t_3) & 1\\ at_3t_1 & a(t_3+t_1) & 1 \end{pmatrix} &= \frac{a^2}{2}\det\begin{pmatrix}t_1t_2 & (t_1+t_2) & 1\\ t_2t_3 & (t_2+t_3) & 1\\ t_3t_1 & (t_3+t_1) & 1 \end{pmatrix} \\ &= \frac{a^2}{2}\det\begin{pmatrix}t_1t_2 & (t_1+t_2) & 1\\ t_2(t_3-t_1) & (t_3-t_1) & 0\\ t_1(t_3-t_2) & (t_3-t_2) & 0 \end{pmatrix} \\ &= \frac{a^2}{2}|(t_2-t_1)(t_3-t_2)(t_1-t_3)| \end{align*} and \begin{align*} \tfrac{1}{2}\det\begin{pmatrix}a(t_1^2+t_1t_2+t_2^2+2) & -at_1t_2(t_1+t_2) & 1\\ a(t_2^2+t_2t_3+t_3^2+2) & -at_2t_3(t_2+t_3) & 1\\ a(t_3^2+t_3t_1+t_1^2+2) & -at_3t_1(t_3+t_1) & 1\\ \end{pmatrix} &= \frac{a^2}{2}\det\begin{pmatrix}(t_1^2+t_1t_2+t_2^2+2) & -t_1t_2(t_1+t_2) & 1\\ (t_2^2+t_2t_3+t_3^2+2) & -t_2t_3(t_2+t_3) & 1\\ (t_3^2+t_3t_1+t_1^2+2) & -t_3t_1(t_3+t_1) & 1\\ \end{pmatrix} \\ &= \frac{a^2}{2}\det\begin{pmatrix}(t_1^2+t_1t_2+t_2^2+2) & -t_1t_2(t_1+t_2) & 1\\ t_3^2-t_1^2+t_2(t_3-t_1) & t_2(t_1^2+t_1t_2-t_2t_3-t_3^2) & 0\\ t_3^2-t_2^2+t_1(t_3-t_2) & t_1(t_2^2+t_2t_1-t_1t_3-t_3^2) & 0\\ \end{pmatrix} \\ &= \frac{a^2}{2}\det\begin{pmatrix}(t_1^2+t_1t_2+t_2^2+2) & -t_1t_2(t_1+t_2) & 1\\ (t_3-t_1)(t_3+t_2+t_1) & t_2(t_1-t_3)(t_1+t_3+t_2) & 0\\ (t_3-t_2)(t_3+t_2+t_1) & t_1(t_2-t_3)(t_1+t_2+t_3)& 0\\ \end{pmatrix} \\ &= \frac{a^2}{2}(t_1+t_2+t_3)^2|(t_2-t_1)(t_3-t_2)(t_1-t_3)| \end{align*} as required. The normals will be concurrent iff the area of their triangle is $0$. This is certainly true if $t_1+t_2+t_3 = 0$. In fact the only if is also true, since no $3$ tangents can be concurrent.

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